A236936 Number T(n,k) of equivalence classes of ways of placing k 9 X 9 tiles in an n X n square under all symmetry operations of the square; irregular triangle T(n,k), n>=9, 0<=k<=floor(n/9)^2, read by rows.

1, 1, 1, 1, 1, 3, 1, 3, 1, 6, 1, 6, 1, 10, 1, 10, 1, 15, 1, 15, 30, 5, 1, 1, 21, 96, 74, 14, 1, 21, 221, 413, 174, 1, 28, 417, 1525, 1234, 1, 28, 705, 4290, 6124, 1, 36, 1107, 10269, 23259, 1, 36, 1638, 21630, 73204, 1, 45, 2334, 41790, 199436

Offset: 9

Author

Christopher Hunt Gribble, Feb 01 2014

Keywords

• tabf
• nonn

Examples

			The first 17 rows of T(n,k) are:
.\ k  0      1      2      3      4
n
9     1      1
10    1      1
11    1      3
12    1      3
13    1      6
14    1      6
15    1     10
16    1     10
17    1     15
18    1     15     30      5      1
19    1     21     96     74     14
20    1     21    221    413    174
21    1     28    417   1525   1234
22    1     28    705   4290   6124
23    1     36   1107  10269  23259
24    1     36   1638  21630  73204
25    1     45   2334  41790 199436
.
T(18,3) = 5 because the number of equivalence classes of ways of placing 3 9 X 9 square tiles in an 18 X 18 square under all symmetry operations of the square is 5.
		

Links

• Christopher Hunt Gribble, C++ program
• Christopher Hunt Gribble, Example graphics

Crossrefs

Cf. A054252, A236679, A236560, A236757, A236800, A236829, A236865, A236915, A236939.

Formula

It appears that:

T(n,0) = 1, n>= 9

T(n,1) = (floor((n-9)/2)+1)*(floor((n-9)/2+2))/2, n >= 9

T(c+2*9,2) = A131474(c+1)*(9-1) + A000217(c+1)*floor(9^2/4) + A014409(c+2), 0 <= c < 9, c even

T(c+2*9,2) = A131474(c+1)*(9-1) + A000217(c+1)*floor((9-1)(9-3)/4) + A014409(c+2), 0 <= c < 9, c odd

T(c+2*9,3) = (c+1)(c+2)/2(2*A002623(c-1)*floor((9-c-1)/2) + A131941(c+1)*floor((9-c)/2)) + S(c+1,3c+2,3), 0 <= c < 9 where

S(c+1,3c+2,3) =

A054252(2,3), c = 0

A236679(5,3), c = 1

A236560(8,3), c = 2

A236757(11,3), c = 3

A236800(14,3), c = 4

A236829(17,3), c = 5

A236865(20,3), c = 6

A236915(23,3), c = 7

A236936(26,3), c = 8